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JEE Main 2024 (8 Apr, Shift 1) · Mass Defect & Binding Energy

Mass Defect & Binding Energy question from JEE Main 2024 (8 Apr, Shift 1)

Q7JEE Main 2024 · 8 APR · SHIFT 1Easy

Mass Defect & Binding Energy

Binding energy of a certain nucleus is 18×108 J18 \times 10^{8}\text{ J}. How much is the difference between total mass of all the nucleons and nuclear mass of the given nucleus:

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Answer: (a)

  1. Mass defect Δm=BEc2=18×1089×1016=2×10−8\Delta m = \dfrac{BE}{c^2} = \dfrac{18 \times 10^8}{9 \times 10^{16}} = 2 \times 10^{-8} kg
  2. 1 μg=10−91\ \mu\text{g} = 10^{-9} kg, so Δm=20 μ\Delta m = 20\ \mug
  3. So the answer is (a) 20 μ\mug.
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